Design of Variable Depth Single-Span Post.pdf
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May 10, 2025
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About This Presentation
Hunched Single Span Bridge: -
(HSSBs) have maximum depth at ends and minimum depth at midspan.
Used for long-span river crossings or highway overpasses when:
Aesthetically pleasing shape is required or
Vertical clearance needs to be maximized
Slide Content
Design of Variable
Depth Single-Span Post
Tensioned Concrete
Bridges (HSSBs)
Preparer : Mohamed Said
Supervisor : KamelFarid
Depth Single-Span Post
Tensioned Concrete
Bridges (HSSBs)
Preparer : Mohamed Said
Supervisor : KamelFarid
Hunched Single Span Bridge
(HSSBs) have maximum depth at ends and minimum depth at
midspan.
Used for long-span river crossings or highway overpasses when:
Aesthetically pleasing shape is required or
Vertical clearance needs to be maximized
(HSSBs) have maximum depth at ends and minimum depth at
midspan.
Used for long-span river crossings or highway overpasses when:
Aesthetically pleasing shape is required or
Vertical clearance needs to be maximized
Simply Supported Bridge
The figure below shows a simply supported bridge having span length Land shot
cantilevers of length a.
Vertical supports carry reaction R.
Maximum positive bending moment occurs at midspan, so it is not possible to
have a shallow section at midspan.
The figure below shows a simply supported bridge having span length Land shot
cantilevers of length a.
Vertical supports carry reaction R.
Maximum positive bending moment occurs at midspan, so it is not possible to
have a shallow section at midspan.
Bridges with Tie-Down
Overview
HSSB is made possible by applying a negative moment to span
ends.
Negative moment is applied by using short cantilevers with tie-
down.
Negative moment at ends results in a reduction in positive moment
at midspanand allows for shallower sections.
HSSB is made possible by applying a negative moment to span
ends.
Negative moment is applied by using short cantilevers with tie-
down.
Negative moment at ends results in a reduction in positive moment
at midspanand allows for shallower sections.
ApplyingadownwardforceFattheendofeachcantileverextensionwill
produceanegativemoment M
-
=FaandincreasethereactionR.
Thetotalnegativemomentisthesumofthemomentduetoself-weightofthe
endcantileverandFa.
Thisnegativemomentreducesthepositivemomentatmidspan.
Principle
ApplyingadownwardforceFattheendofeachcantileverextensionwill
produceanegativemoment M
-
=FaandincreasethereactionR.
Thetotalnegativemomentisthesumofthemomentduetoself-weightofthe
endcantileverandFa.
Thisnegativemomentreducesthepositivemomentatmidspan.
Principle
ThevaluesofFandacanbeadjustedtoobtainanydesireddistributionof
positiveandnegativemoment.
Theparameteramustbeselectedtominimizethetotalstructurelength,while
keepingthedesignforshearintheendcantileversmanageable.
Principle
ThevaluesofFandacanbeadjustedtoobtainanydesireddistributionof
positiveandnegativemoment.
Theparameteramustbeselectedtominimizethetotalstructurelength,while
keepingthedesignforshearintheendcantileversmanageable.
Principle
Itisreasonabletoprescribeadownwardtie-downforceor“action”tohavethe
requireddistributionofpositiveandnegativemoments.
Thebridgethenisasimplysupportedstaticallydeterminatebridgeastheforce
F isknown.
Thiswillhelptoavoidunfavourablesecondarymoments(duetoprestressing,
creep,andshrinkage)thatcomplicatethedesign.
Principle
Itisreasonabletoprescribeadownwardtie-downforceor“action”tohavethe
requireddistributionofpositiveandnegativemoments.
Thebridgethenisasimplysupportedstaticallydeterminatebridgeastheforce
F isknown.
Thiswillhelptoavoidunfavourablesecondarymoments(duetoprestressing,
creep,andshrinkage)thatcomplicatethedesign.
Principle
DESIGN EXAMPLE
Main Span = 270 ft
Cantilever length a = Span/6 = 45 ft
Tie down location 2m from ends
Overall length = 364 ft
Main Span = 270 ft
Cantilever length a = Span/6 = 45 ft
Tie down location 2m from ends
Overall length = 364 ft
DESIGN EXAMPLE
DESIGN EXAMPLE
Two-cell/three-web box girder
Two 12ft traffic lanes and two 10 ftshoulders
Clear roadway 44 ft, overall width of 47 ft.
Span/depth ration = 40 (Abutment) & (20 at Midspan)
depth of girder =
13’6” at abutment
6’9” at midspan
Web thickness = 16”
Top slab th.= 9”
Bottom slab th.= 8” at midspan& 14 at abutment
Two-cell/three-web box girder
Two 12ft traffic lanes and two 10 ftshoulders
Clear roadway 44 ft, overall width of 47 ft.
Span/depth ration = 40 (Abutment) & (20 at Midspan)
depth of girder =
13’6” at abutment
6’9” at midspan
Web thickness = 16”
Top slab th.= 9”
Bottom slab th.= 8” at midspan& 14 at abutment
DESIGN EXAMPLE : Loads
Self-weight density of concrete = 155 lb/ft
3
Superimposed dead load (barrier + ws) = 1852 lb/ft
Live load according to AASHTO:
HL-93
four lanes
multiple presence factor = 0.65
dynamic load allowance = 1.33
Fc’ = 6000 psi
Self-weight density of concrete = 155 lb/ft
3
Superimposed dead load (barrier + ws) = 1852 lb/ft
Live load according to AASHTO:
HL-93
four lanes
multiple presence factor = 0.65
dynamic load allowance = 1.33
Fc’ = 6000 psi
Section Properties
Property Midspan Abutment
A, ft
2
84.28 120.65
I, ft
4
556.3 3310.3
y
t
, ft 2.690 6.093
y
b
, ft 4.060 7.407
r
2
, ft
2
6.601 27.436
r
2
/y
t
, ft 2.454 2.503
r
2
/y
b
, ft 1.626 3.704
Q, ft
3
------ 292.814
Property Midspan Abutment
A, ft
2
84.28 120.65
I, ft
4
556.3 3310.3
y
t
, ft 2.690 6.093
y
b
, ft 4.060 7.407
r
2
, ft
2
6.601 27.436
r
2
/y
t
, ft 2.454 2.503
r
2
/y
b
, ft 1.626 3.704
Q, ft
3
------ 292.814
Bending Moments and Shear
Moment Midspan Abutment
M
SW
, ft.kip 108,700 -20,700
M
SDL
, ft.kip 14,800 -2000
M
LL
, ft.kip 31,000 -11,000
M
TOTAL
, ft.kip 154,500 -33,700Shear Outside Abutment Inside Abutment
V
SW
, kip 879 2050
V
SDL
, kip 87 250
V
LL
, kip 327 471
V
TOTAL
, kip 1293 2271
y
b
Where:
SW: self-weight
SDL: superimposed dead laoad
LL: live load
TD: tie-down
Moment Midspan Abutment
M
SW
, ft.kip 108,700 -20,700
M
SDL
, ft.kip 14,800 -2000
M
LL
, ft.kip 31,000 -11,000
M
TOTAL
, ft.kip 154,500 -33,700Shear Outside Abutment Inside Abutment
V
SW
, kip 879 2050
V
SDL
, kip 87 250
V
LL
, kip 327 471
V
TOTAL
, kip 1293 2271
y
b
Where:
SW: self-weight
SDL: superimposed dead laoad
LL: live load
TD: tie-down
Analysis
Assuming tie-down force = 2276 kip.
Lever arm = 45 ft.
Moment due to tie-down = 2276*45 =102,400 ft.kip.
Tie-down moment shifts down positive moment from 108,700 to 6,300 ft.kip.
Tie-down moment shifts up negative momentfrom 20,700 to 123,100 ft.kip.
Small reserve of positive moment at midspanis desirable to avoid stress reversal
when live load acts on the bridge.
Assuming tie-down force = 2276 kip.
Lever arm = 45 ft.
Moment due to tie-down = 2276*45 =102,400 ft.kip.
Tie-down moment shifts down positive moment from 108,700 to 6,300 ft.kip.
Tie-down moment shifts up negative momentfrom 20,700 to 123,100 ft.kip.
Small reserve of positive moment at midspanis desirable to avoid stress reversal
when live load acts on the bridge.
The shear value due to SW and TD is :
3155 kip on the cantilever side of the abutment (outside abutment).
2050 kip on the main span side of the abutment (inside abutment).
The tie down force has no effect on the main span shear.
The designer can select the tie-down force to shift the SW moment diagram by
any amount
The method used here is to lay out a reasonable number of continuity
prestressingtendons
The chosen number of tendons used in this example is based on zero tension
under dead load and live load, at the bottom at midspanand at the top of the
abutments.
The shear value due to SW and TD is :
3155 kip on the cantilever side of the abutment (outside abutment).
2050 kip on the main span side of the abutment (inside abutment).
The tie down force has no effect on the main span shear.
The designer can select the tie-down force to shift the SW moment diagram by
any amount
The method used here is to lay out a reasonable number of continuity
prestressingtendons
The chosen number of tendons used in this example is based on zero tension
under dead load and live load, at the bottom at midspanand at the top of the
abutments.
Design Procedure
Design Procedure steps
1.
Lay out continuity tendons (at midspanand abutments).
2.
Determine the positive moment that these tendons can carry.
3.
Calculate tie down moment (difference between tendons moment and total
moment at midspan).
4.
Calculate total negative moment at abutments.
5.
Determine the prestressingrequired at the abutments.
6.
Determine the additional local top tendons required at the abutments to satisfy
value at step 5.
7.
Check principal stresses at the abutments (on both sides).
8.
Check flexural strength (at midspanand the abutments).
9.
Plot flexural stress diagrams and check flexural stresses.
1.
Lay out continuity tendons (at midspanand abutments).
2.
Determine the positive moment that these tendons can carry.
3.
Calculate tie down moment (difference between tendons moment and total
moment at midspan).
4.
Calculate total negative moment at abutments.
5.
Determine the prestressingrequired at the abutments.
6.
Determine the additional local top tendons required at the abutments to satisfy
value at step 5.
7.
Check principal stresses at the abutments (on both sides).
8.
Check flexural strength (at midspanand the abutments).
9.
Plot flexural stress diagrams and check flexural stresses.
Step1: Tendons layout
Six-19 strand continuity tendons per web is selected ( Total tendons
= 3x6=18)
Tendons are of grade 270, low-relaxation, 0.6 in. diameter seven
wire steel strand.
The tendons are laid out as shown in the figure below
Tendons travel the full length of the bridge, with low point at
midspan, high points at the abutment, and anchorages near the
tie down locations.
Six-19 strand continuity tendons per web is selected ( Total tendons
= 3x6=18)
Tendons are of grade 270, low-relaxation, 0.6 in. diameter seven
wire steel strand.
The tendons are laid out as shown in the figure below
Tendons travel the full length of the bridge, with low point at
midspan, high points at the abutment, and anchorages near the
tie down locations.
Step 1: Tendons layout
The following additional details show that the centerof gravity of
the tendons is 16.25 in. from both the top and bottom of the section
At the anchorage zone the webs are flared from 16 to 24 in. over a
length of 10 ftto accommodate the anchorages, then the center
of gravity of the tendons is 52 in. from the top
The following additional details show that the centerof gravity of
the tendons is 16.25 in. from both the top and bottom of the section
At the anchorage zone the webs are flared from 16 to 24 in. over a
length of 10 ftto accommodate the anchorages, then the center
of gravity of the tendons is 52 in. from the top
Step1: Additional Prestressing
details
details
Step1: Tendons layout
Step1: Tendons layout
Step 1: Tendons layout
The preliminary design is based on jacking stress of 0.75 f
pu,
an initial stress
of 0.7f
pu
after short-term losses, and a final effective stress of 0.6 f
pu
For this tendon layout the effective prestressingforce is P
P = 0.6 x 270 ksix 0.217 in2 = 35.15 kip per strand
Total force = 35.15 x 342 = 12,023 kip (six tendons/web totalling 342
strands )
The eccentricity at bot. e
bot
= y
b
-y
bar
= 4.0598-1.3542 = 2.7056 ft
The eccentricity at top. e
top
= y
t
-y
bar
= 6.0925-1.3542 = 4.7384 ft
The preliminary design is based on jacking stress of 0.75 f
pu,
an initial stress
of 0.7f
pu
after short-term losses, and a final effective stress of 0.6 f
pu
For this tendon layout the effective prestressingforce is P
P = 0.6 x 270 ksix 0.217 in2 = 35.15 kip per strand
Total force = 35.15 x 342 = 12,023 kip (six tendons/web totalling 342
strands )
The eccentricity at bot. e
bot
= y
b
-y
bar
= 4.0598-1.3542 = 2.7056 ft
The eccentricity at top. e
top
= y
t
-y
bar
= 6.0925-1.3542 = 4.7384 ft
Step 2: Determine Positive moment
carried by the continuity tendons
M
+
= P (e
bot
+ r
2
/y
b
) = 12,023 x (2.7056+1.6258) = 52,100 ft.kip
carried by the continuity tendons
M
+
= P (e
bot
+ r
2
/y
b
) = 12,023 x (2.7056+1.6258) = 52,100 ft.kip
Step 3: Calculate tie-down
moment
Difference between M
TD
= 52,100 –154,500 = -102,400 ft.kip
moment
Difference between M
TD
= 52,100 –154,500 = -102,400 ft.kip
Step 4: Calculate the total
negative moment at the abutment
Total negative moment = M
TD
+ total abutment moment
M
-
= -102,400 + -33,700 = -136,100 ft.kip
negative moment at the abutment
Total negative moment = M
TD
+ total abutment moment
M
-
= -102,400 + -33,700 = -136,100 ft.kip
Step 5: Determine the prestressing
required at the abutment
P
top
= M
-
/(e
top
+ r
2
/y
t
) = 136,100 x (4.7381+4.5033) = 14,726 kip
required at the abutment
P
top
= M
-
/(e
top
+ r
2
/y
t
) = 136,100 x (4.7381+4.5033) = 14,726 kip
Step 6: Determine the additional local
top tendons required at abutment
P
additoinal
= 14,726 -12,023 = 2703 kip
Number of strands = 2703 / 35.15 = 76.9 ( use 76 )
The additional prestressingcan be provided by four 19-strand
tendon at the top of the section over the abutment.
A tendon length of 45 ftwill be adequate, with the added tendons
anchored in small buildouts at the web/top-slab interface , located
22’ 6” on each side of the abutment.
top tendons required at abutment
P
additoinal
= 14,726 -12,023 = 2703 kip
Number of strands = 2703 / 35.15 = 76.9 ( use 76 )
The additional prestressingcan be provided by four 19-strand
tendon at the top of the section over the abutment.
A tendon length of 45 ftwill be adequate, with the added tendons
anchored in small buildouts at the web/top-slab interface , located
22’ 6” on each side of the abutment.
Step 7: Check principal stresses
The calculations for the principal stress check are summarized in the
next table. These calculations are made for both sides of the
abumtnet.
The flexure stress σ
x
= P/A , and the corresponding shear τ
xy
= VQ/Ib
are used
Considering outside the abutment first, the axial force P = 14726 kip is the
prestressingrequired at the abutment, and the shear force V=3569 kip is the
combined shear due to loads and the tie down shear
The calculations for the principal stress check are summarized in the
next table. These calculations are made for both sides of the
abumtnet.
The flexure stress σ
x
= P/A , and the corresponding shear τ
xy
= VQ/Ib
are used
Considering outside the abutment first, the axial force P = 14726 kip is the
prestressingrequired at the abutment, and the shear force V=3569 kip is the
combined shear due to loads and the tie down shear
Step 7: Check principal stresses
The flexural stresses and shear stress are determined as : σ
x
= -848 psi , and τ
xy
= 548
psi.
These stresses are used to calculate radius R=693 psi for the Mohr’s circle diagram
, which indicates a maximum principal tension of 269 psi.
This stress is just under the allowable value of 3.5 ????????????′
= 271 psi
This means adequate section and it is possible to reinforce this section without
the need to increase the section depth of web thickness.
For section inside the abutment same steps are followed and summarized in the
next table.
The flexural stresses and shear stress are determined as : σ
x
= -848 psi , and τ
xy
= 548
psi.
These stresses are used to calculate radius R=693 psi for the Mohr’s circle diagram
, which indicates a maximum principal tension of 269 psi.
This stress is just under the allowable value of 3.5 ????????????′
= 271 psi
This means adequate section and it is possible to reinforce this section without
the need to increase the section depth of web thickness.
For section inside the abutment same steps are followed and summarized in the
next table.
Check on principal stresses
Parameter
Principal stress
Six tendons/web
Outside Abutment Inside Abutment
P, kip 14.726 14.726
V
loads
, kip 1293 1293
V
TD
, kip 2276 0
V,kip 3569 2771
σ
x
, psi -848 -848
τ
xy
, psi 548 425
R, psi 693 601
σ
min
, psi -1117 -1024
σ
max
, psi 269 177
<3.5 ????????????′
271 271
Parameter
Principal stress
Six tendons/web
Outside Abutment Inside Abutment
P, kip 14.726 14.726
V
loads
, kip 1293 1293
V
TD
, kip 2276 0
V,kip 3569 2771
σ
x
, psi -848 -848
τ
xy
, psi 548 425
R, psi 693 601
σ
min
, psi -1117 -1024
σ
max
, psi 269 177
<3.5 ????????????′
271 271
Step 8: Check flexural strength
Calculations for the strength check are summarized in the next table
Calculations are made both at midspanand the abutment.
The demand-capacity ratio (D/C) is 0.867 at midspanand 0.753 at
the abutment.
These values indicate that the design is adequate with respect to
flexural strength
Calculations for the strength check are summarized in the next table
Calculations are made both at midspanand the abutment.
The demand-capacity ratio (D/C) is 0.867 at midspanand 0.753 at
the abutment.
These values indicate that the design is adequate with respect to
flexural strength
Check on flexural strength
Parameter
flexural strength
Six tendons/web
Outside Abutment Inside Abutment
M
u
, ft.kip 80,600 -175,600
M
cr
, ft.kip 75,600 -222,400
ϕM
n
, ft.kip 80,600 -222,400
M
r
, ft.kip 93,000 -295,300
D/C 0.867 0.753
Parameter
flexural strength
Six tendons/web
Outside Abutment Inside Abutment
M
u
, ft.kip 80,600 -175,600
M
cr
, ft.kip 75,600 -222,400
ϕM
n
, ft.kip 80,600 -222,400
M
r
, ft.kip 93,000 -295,300
D/C 0.867 0.753
Step 9: Check flexural stresses
Flexural stresses over the full length are calculated and checked
The basis for this design (0 psi tension at the bottom midspanunder combined
dead and live load) is satisfied.
The corresponding stress at the top of the section at the abutments is 319 psi
tension. The added local top tendons will counteract and reduce this stress to 0
psi.
The maximum initial compressive stress at the bottom is -2488 psi, which is less
than the initial allowable compressive stress of 0.6 f
ci
’
= 0.6 x 4500 = 2700 psi.
This is also the maximum compressive stress, so all the compressive stresses are
less than the final allowable compressive stress of 0.45 x f
c
’ = 0.45 x 6000 psi = 2700
psi.
Flexural stresses over the full length are calculated and checked
The basis for this design (0 psi tension at the bottom midspanunder combined
dead and live load) is satisfied.
The corresponding stress at the top of the section at the abutments is 319 psi
tension. The added local top tendons will counteract and reduce this stress to 0
psi.
The maximum initial compressive stress at the bottom is -2488 psi, which is less
than the initial allowable compressive stress of 0.6 f
ci
’
= 0.6 x 4500 = 2700 psi.
This is also the maximum compressive stress, so all the compressive stresses are
less than the final allowable compressive stress of 0.45 x f
c
’ = 0.45 x 6000 psi = 2700
psi.
Abutment Details
•The box girders have 2 ft
thick diaphragms at the
abutments and tie-down
locations .
•The abutment diphrams
transfer the bearing
reactions from the web,
while the tie-down
diaphramstransfer the
tie-down forces to the
web .
•The box girders have 2 ft
thick diaphragms at the
abutments and tie-down
locations .
•The abutment diphrams
transfer the bearing
reactions from the web,
while the tie-down
diaphramstransfer the
tie-down forces to the
web .
Abutment Details
•Thetie-downsconsistsof
24rock/soilanchors.
•Eachanchorconsistsof
five0.6in.diameter
grade270steelstrands.
•Theanchorsareplaced
intworowsandare
spacedat2ftcenters
acrossthewidthofthe
boxgirder.
•Simpleabutments
•Thetie-downsconsistsof
24rock/soilanchors.
•Eachanchorconsistsof
five0.6in.diameter
grade270steelstrands.
•Theanchorsareplaced
intworowsandare
spacedat2ftcenters
acrossthewidthofthe
boxgirder.
•Simpleabutments
Abutment Details
•Eachanchorhasa
stressingheadatthetop
andadevicethat
trasfersforcetotherock
orsoilattheembedded
end
•Thebridgeissupported
bypotbearinglocated
ontopofthepilecaps
•Thesebearingcarrytotal
loadreaction+tiedown
force.
•Eachanchorhasa
stressingheadatthetop
andadevicethat
trasfersforcetotherock
orsoilattheembedded
end
•Thebridgeissupported
bypotbearinglocated
ontopofthepilecaps
•Thesebearingcarrytotal
loadreaction+tiedown
force.
Abutment Alternatives
Simple abutments have been shown for this design
example
However, wide varaietyof substructure alternatives may
be used.
One maybe raising the bridge to have regular
abutments with wingwalls.
Other could be supporting the bridge on piers at the
location of bearings and having abutments at the
location of tie-downs
Simple abutments have been shown for this design
example
However, wide varaietyof substructure alternatives may
be used.
One maybe raising the bridge to have regular
abutments with wingwalls.
Other could be supporting the bridge on piers at the
location of bearings and having abutments at the
location of tie-downs
Conclusions
This presentation provides simple procedure for the design of
variable depth single-span post-tensioned concrete box girder
bridge
This allows aesthetically pleasing bridges to be designed and
constructed
The HSSB design method is relatively simple, and even novice
bridge designers can create very beautiful bridges, as these
bridges are no more difficult to design than simply supported single
span bridges.
This HSSB have all the benefits of continuous bridge, without any of
the complexities involved in their design.
This presentation provides simple procedure for the design of
variable depth single-span post-tensioned concrete box girder
bridge
This allows aesthetically pleasing bridges to be designed and
constructed
The HSSB design method is relatively simple, and even novice
bridge designers can create very beautiful bridges, as these
bridges are no more difficult to design than simply supported single
span bridges.
This HSSB have all the benefits of continuous bridge, without any of
the complexities involved in their design.
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